CAIE P2 2008 June — Question 1 3 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2008
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInequalities
TypeSolve absolute value inequality
DifficultyEasy -1.2 This is a straightforward absolute value inequality requiring only the standard technique of splitting into two cases: -2 < 3x - 1 < 2, then solving linear inequalities. It's a routine textbook exercise with minimal steps and no conceptual challenge beyond recalling the definition of absolute value.
Spec1.02l Modulus function: notation, relations, equations and inequalities

1 Solve the inequality \(| 3 x - 1 | < 2\).

AnswerMarks
State or imply non-modular inequality \((3x - 1)^2 < 2^2\), or corresponding equation or pair of linear equationsM1
Obtain critical values \(-\frac{1}{3}\) and \(1\)A1
State correct answer \(-\frac{1}{3} < x < 1\)A1
[3]
OR
AnswerMarks
State one critical value, e.g. \(x = 1\), by solving a linear equation (or inequality) or from a graphical method or by inspectionB1
State the other critical value correctlyB1
State correct answer \(-\frac{1}{3} < x < 1\)B1
[3]
State or imply non-modular inequality $(3x - 1)^2 < 2^2$, or corresponding equation or pair of linear equations | M1 |
Obtain critical values $-\frac{1}{3}$ and $1$ | A1 |
State correct answer $-\frac{1}{3} < x < 1$ | A1 |
| [3] |

**OR**

State one critical value, e.g. $x = 1$, by solving a linear equation (or inequality) or from a graphical method or by inspection | B1 |
State the other critical value correctly | B1 |
State correct answer $-\frac{1}{3} < x < 1$ | B1 |
| [3] |
1 Solve the inequality $| 3 x - 1 | < 2$.

\hfill \mbox{\textit{CAIE P2 2008 Q1 [3]}}